From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Problem
Date: 23 Jul 1996 04:49:52 GMT

In article <4t0um4$3r5@news1.t1.usa.pipeline.com>,
 <mchen@nyc.pipeline.com> wrote:
> 
>Find all functions f, from the positive reals to themselves, which satisfy
>xy f(x+y) [f(x)+f(y)] = 1 for all positive x, y. 


What on earth would prompt such a question? Oh well, the only such  f
is the reciprocal function  (f(x) = 1/x). While there are undoubtedly
quick ways to see this once you know what the answer is, let me show
you what I do to answer such questions.

Given a relation which holds for all  x  and  y  we try to see simpler
forms which result from particular combinations of  x  and  y. In this
problem x=0 and y=0 are forbidden, and  x=1 (say) does not appear
helpful, but taking  y=x  gives a doubling formula
	f(2x) = 1/[2 x^2 f(x) ]
which can be iterated to get a nicer quadrupling formula
	f(4x) = f(x)/4.

(You can use this to at least establish a sort of periodicity:
	k(x) = 1/[ exp(x) f(exp(x)) ]
is periodic with period  ln(4). In fact,  we will show  k  is constant.)

Another perspective is obtained by considering the given relation as an
additivity formula:
	f(x+y) = 1/[ xy ( f(x) + f(y) ) ]
This is useful since addition on the positive reals is associative.
(It's commutative too, but since this formula is already symmetric in 
x  and  y  we get no further information that way.) Use associativity
to set  f( x + (y+z) ) = f( (x+y) + z ); repeated application of the
additivity formula "reduces" this to a terrible equation in  x, y, z,
f(x), f(y), and f(z). The advantage to this is that if  5  of these are
held fixed, the last may be computed; in fact, there's a quadratic
polynomial to solve which allows us to compute  f(y)  as a function of
y  (involving a few other numbers which we can fix as constants).

We can combine the two ideas by fixing  x=1,  z=4,  f(x) = a (say), so
that  f(z) = a/4; then  f(y)  becomes a rather explicit function involving
one unknown constant  a.

We attempt to deduce the value of  a  by plugging in one nice value for  y.
For example, if  y  is also equal to  4, then we compare the two expansions
for  f(1+(4+4)) = f((1+4)+4), each of which is a rational expression in
te one unknown  a = f(1), and after some simplification we get
a^2=1. Since  a  is positive, we must have  a=1. We may then simplify the
expression of the preceding paragraph and after letting the fur fly for
a while, we finally wrestle the formula down to a simple  f(y) = 1/y. 

(If the positivity condition is relaxed, then this analysis shows that
f(y) = -1/y  is another solution.)

dave

By the way, it's often easier to guess the solution if we're told the
context in which the problem arises. (If there is no context, then
maybe the problem is best left out of sight.)

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